Simplyifying a Boolean Expression that has three variables in it

Question

The following problem can be found at the following URL: Some problems related to digital logic

It is problem number 26 on the webpage.
Problem:

Simplify the following expression using Boolean Algebra: $$ (X + Z)(\overline X + Y)(Z + Y) $$
Answer: \begin{align*} (X + Z)(\overline X + Y)(Z + Y) &= ( 0 + Z \overline X + XY + ZY )(Z + Y) \\ (X + Z)(\overline X + Y)(Z + Y) &= ( Z \overline X + XY + ZY )(Z + Y) \\ (X + Z)(\overline X + Y)(Z + Y) &= Z( Z \overline X + XY + ZY ) + Y( Z \overline X + XY + ZY ) \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X + XYZ + ZY + Y( Z \overline X + XY + ZY ) \\ % (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X + XYZ + ZY + YZ \overline X + XY + ZY \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X + XYZ + ZY + XY + ZY \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X + ZY + XY \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X + YZ + XY \\ \end{align*} However, the author of the problem has the following answer: $$ XY + \overline X Z $$ Where did I go wrong?

Answer 1

Just continue from your last line $$ (X + Z)(\overline X + Y)(Z + Y) = Z \overline X + YZ + XY $$ and notice that $$ \begin{align} (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X (Y + \overline Y) + Z(X + \overline X)Y + (Z + \overline Z)XY \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X (\color{blue} Y + \overline Y) + Z(\color{red}X + \color{blue} {\overline X})Y + (\color{red} Z + \overline Z)XY \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X (\color{blue} Y + \overline Y) + (\color{red} Z + \overline Z)XY \\ (X + Z)(\overline X + Y)(Z + Y) &= Z \overline X + XY. \\ \end{align} $$

Answer 2

You followed a good process. But perhaps the easiest way to see what you missed is to notice this:

$$\begin{align*} F &= Z\overline{X}+YZ+XY\\ &=Z\overline{X}+XY+YZ\\ &= \underbrace{\left(Z\overline{X}+XY\right)}+ \,YZ \end{align*}$$

The gathered up part (the left term of the last equation above) is just a mux:

^{simulate this circuit – Schematic created using CircuitLab}

If you feed in \$Z=1\$ and \$Y=1\$ then the mux output has to be 1. If you or into your equation \$+Y Z\$, this contributes nothing when that term is true because the only way that term is true is if \$Z=1\$ and \$Y=1\$. But the mux already yields that term, in that case. So the right term of the last equation above makes no contribution.